# Video: Pack 3 • Paper 1 • Question 21

Pack 3 • Paper 1 • Question 21

07:06

### Video Transcript

The speed-time graph shows the first 55 seconds of Michael’s journey on his scooter. Michael travelled a total distance of 650 meters in the 55 seconds. Work out the value of 𝑣.

There are two things we can work out from this speed-time graph aside from the speed at any given point of the journey: the first is the acceleration that’s given by the gradient of the graph at a given point; the second is the total distance travelled, which is given by the area under the curve, that’s the area between the graph and the 𝑥-axis.

This part of the question is asking us to find the value of 𝑣. However, it also tells us the total distance travelled, which means we’ll need to find the area under the graph. We can begin by splitting the shape up into parts. We can either choose three parts or possibly four, depending on how confident you are finding the area of a trapezium. Let’s stick with three for now.

Between zero and 20 seconds, we have a triangle. The formula for area of a triangle is a half multiplied by its base multiplied by its height. The width of the base of this triangle is 20 and its height is 𝑣. The formula then becomes a half multiplied by 20 multiplied by 𝑣. Since a half of 20 is 10, the area of the first part of this graph is 10𝑣.

The third shape is also a triangle. So let’s calculate the area of the third shape now. The width of the base of this triangle is calculated by subtracting 40 from 55. That gives us 15. Its height is 𝑣 plus five. So the area of this triangle is a half multiplied by 15 multiplied by 𝑣 plus five. A half of 15 is a decimal. So instead, we’ll expand the brackets. 15 multiplied by 𝑣 is 15𝑣 and 15 multiplied by five is 75. That means that the area of the third shape is 15𝑣 plus 75 all over two.

The second shape is a trapezium. The formula for area of a trapezium is a half multiplied by 𝑎 plus 𝑏 or multiplied by ℎ, where 𝑎 and 𝑏 are the lengths of the parallel sides of the trapezium and ℎ is the height between them. The lengths of the parallel sides of this trapezium are 𝑣 and 𝑣 plus five and the height between them is 20 that’s found by subtracting 20 from 40.

We can simplify the brackets by collecting like terms to give us two 𝑣 plus five. And since multiplication is commutative, that means you can do it in any order. We can move the 20 next to the half. A half of 20 is 10. Then, finally, we can expand the brackets. 10 multiplied by two 𝑣 is 20𝑣 and 10 multiplied by five is 50. The area of the trapezium is 20𝑣 plus 50.

Remember we said that the total area under the graph represents the distance travelled. In our case, that’s 650 meters. So we can add together our expressions and make them equal to 650 to find an equation in terms of 𝑣. Now, this fraction is making our life a little tricky. So we’ll multiply everything in this equation by two to get rid of the two on the denominator. That gives us 1300 is equal to 20𝑣 plus 40𝑣 plus 100 plus 15𝑣 plus 75.

We can then collect together like terms to give us 1300 is equal to 75𝑣 plus 175. Next, we’ll subtract 175 from both sides of the equation, giving us 1125 is equal to 75𝑣. Finally, we need to divide both sides of the equation by 75.

We can use long division to help us perform that calculation. One 75 is 75 and that’s the most number of 75s we can use to make 112. We then subtract 75 from 112, which gives us 37. We bring down this remaining five and work out how many 75s we need to make 375. Five 75s are 375. And then, once again, we subtract 375 from 375, giving us zero. That means 1125 divided by 75 is 15 with no remainder. 𝑣 is equal to 15.

Describe Michael’s acceleration for each part of the 55-second journey.

Before we do anything, let’s change the value of 𝑣 on our diagram. 𝑣 was 15. Therefore, 𝑣 plus five is 20. The two values on our 𝑦-axis then representing speed are 15 and 20. Now, remember we said that the gradient of the graph tells us the acceleration at a given point. We can therefore find the gradient of each part of the journey using the formula gradient is change in 𝑦 over change in 𝑥. That sometimes called rise over run.

Let’s consider the first part of the journey. The change in 𝑦 is the rise. It’s the height of that triangle, which is 15. The change in 𝑥 is the run. It’s the width of the triangle, which we worked out earlier to be 20. The gradient is therefore 15 over 20, which we can simplify to three-quarters by dividing by the numerator and the denominator by five.

Now, let’s consider the second part of the journey. The height of this triangle which represents the change in 𝑦 is five. It’s the difference between 20 and 15. The width of the triangle which represents the change in 𝑥 is 20. So the gradient is five over 20, which once again we can simplify by dividing both the numerator and the denominator by five. The gradient during the second part of the journey is a quarter.

Finally, let’s consider the third part of our journey. This part of the graph is sloping down. That means it has a negative gradient and represents a deceleration. The change in 𝑦 is therefore negative 20, given by the height of the triangle. The change in 𝑥 is 15. So the gradient of this part of the journey is negative 20 over 15. That simplifies to negative four over three.

We mustn’t forget that when writing the gradient as an acceleration, we need to include units — in this case that’s metres per second squared. For the first 20 seconds, he is accelerating at a constant rate of three-quarters metres per second squared. For the next 20 seconds, he is accelerating at a constant rate of a quarter metres per second squared. For the final 15 seconds, he is decelerating at a constant rate of four-thirds metres per second squared.